A3 - SURDS

A3: SURDS ● Four operations on surds, including rationalising the denominator ● Solving equations involving surds 1. ABCD is a trapezium where AB is parallel to DC. The area of the trapezium is . 12 + 11 3 𝑐 𝑚 2 cm and cm. 𝐴𝐵 = 8 − 48 𝐷𝐶 = 12 − 12 Find the perpendicular distance between and , leaving your answer 𝐴𝐵 𝐷𝐶 in the form , where and are integers. 𝑎 + 𝑏 3 𝑎 𝑏 [5] 2. A rectangular block has a square base of side cm and a height of h 10+ 2 ( ) cm. The volume of the rectangular block is . 52 + 28 5 ( ) 𝑐 𝑚 3 Without using a calculator , find the height of the rectangular block, in cm. Give your answer in the form cm where and are integers. 𝑎 + 𝑏 5 ( ) 𝑎 𝑏 [5] 3. It is given that , where and are integers. 𝑝 − 𝑞 5 = 12 3 + 5 𝑝 𝑞 Without using a calculator, find the value of and . 𝑝 𝑞 [6] 4. Simplify . 7 2− 3 8+ 32 [2] 5. A rectangle has an area of . Given that the length is cm, 11 − 7 𝑐 𝑚 2 3 + 7 find the breadth of the rectangle and express the breadth in the form , 𝑎 − 𝑏 7 where and are integers. 𝑎 𝑏 [4] 6. It is given that . Find in the form , 2 3( 𝑥 + 1 ) = 2 − 𝑥 𝑥 𝑎 3− 𝑏 where and are real numbers. 𝑎 𝑏 [4] 7. Show can be written in the form , where and 2 5− 3 ( ) 6 + 5 ( )3 + 5 𝑎 + 𝑏 5 𝑎 𝑏 are rational numbers. [4] 8. Solve the equation , where , giving your answer in its 𝑥 3 − 4 𝑥 = 3 𝑥 ≠ 0 simplest surd form. [4]

9. (a) Without using a calculator, express in the form , 2 3+ 2 ( ) 2 𝑎 + 𝑏 6 where and are integers. 𝑎 𝑏 [2] 9. (b) Hence, or otherwise, express in the form , 2 3+ 2 ( ) 2 3 + 6 𝑝 + 𝑞 6 where and are rational numbers. 𝑝 𝑞 [3] 10. It is given that . Find in the form , where 3( 𝑥 + 3 ) = 𝑥 + 17 𝑥 𝑎 + 𝑏 3 where and are integers. 𝑎 𝑏 [4] 11. Given that , find the values of the integers 𝑎 − 3 5 ( ) 2 + 5 ( )= 𝑏 − 2 5 and . 𝑎 𝑏 [3] 12. Solve , leaving your answer in the form , 𝑥 + 7− 3 2= 1 𝑚 + 𝑛 2 where and are integers. 𝑚 𝑛 [3] 13. Simplify , giving your answer in the form , 108 − 12 3 𝑘 3 where is an integer. 𝑘 [2] 14. Without using a calculator, solve the equation . 3 𝑥 − 2= 2 1 + 𝑥 2 ( ) [5]

A3: SURDS (MARKING SCHEME) 1. ABCD is a trapezium where AB is parallel to DC. The area of the trapezium is . 12 + 11 3 𝑐 𝑚 2 cm and cm. 𝐴𝐵 = 8 − 48 𝐷𝐶 = 12 − 12 Find the perpendicular distance between and , leaving your answer 𝐴𝐵 𝐷𝐶 in the form , where and are integers. 𝑎 + 𝑏 3 𝑎 𝑏 Area of trapezium = , where and are the length of parallel sides 1 2 ( 𝑎 + 𝑏 ) ℎ 𝑎 𝑏 12 + 11 3 = 1 2 8 − 48+ 12 − 12 ( )ℎ ★ 24 + 22 3= 20 − 4 × 12− 12 ( )ℎ 48= 4 × 12 ★ 24 + 22 3= 20 − 2 12− 12 ( )ℎ 2 2 × 12= 2 12 24 + 22 3= 20 − 3 12 ( )ℎ 24 + 22 3= 20 − 3 4 × 3 ( )ℎ ★ 24 + 22 3= 20 − 6 3 ( )ℎ 3 2 2 × 3= 3 ( 2 ) 3 24 + 22 320 − 6 3 = ℎ ℎ = 24 + 22 320 − 6 3 × 20 + 6 320 + 6 3 ★ Rationalise the denominator ℎ = 12 + 11 310 − 3 3 × 10 + 3 310 + 3 3 ℎ = 120 + 36 3+ 110 3+ 99 100 − 27 ℎ = 219 + 146 373 ℎ = 3 + 2 3 cm ℎ = 3 + 2 3 [5]

2. A rectangular block has a square base of side cm and a height of h 10+ 2 ( ) cm. The volume of the rectangular block is . 52 + 28 5 ( ) 𝑐 𝑚 3 Without using a calculator , find the height of the rectangular block, in cm. Give your answer in the form cm where and are integers. 𝑎 + 𝑏 5 ( ) 𝑎 𝑏 Volume of rectangular block = 𝑙𝑏ℎ 52 + 28 5 ( )= 10+ 2 ( ) 2 ℎ ℎ = 52 + 28 5 10+ 2 ( ) 2 ℎ = 52 + 28 510 + 2 20+ 2 ★ ℎ = 52 + 28 512 + 2 4 × 5 2 20= 2 4 × 5 ℎ = 52 + 28 512 + 4 5 ★ Rationalise the denominator ℎ = 52 + 28 512 + 4 5 × 12 − 4 512 − 4 5 ℎ = 624 − 208 5+ 336 5− 112 ( 5 ) 144 − 16 ( 5 ) ℎ = 64 + 128 564 ℎ = 1 + 2 5 cm ℎ = 1 + 2 5 [5]

3. It is given that , where and are integers. 𝑝 − 𝑞 5 = 12 3 + 5 𝑝 𝑞 Without using a calculator, find the value of and . 𝑝 𝑞 𝑝 − 𝑞 5 = 12 3 + 5 𝑝 − 𝑞 5 = 12 3 + 5 ( ) 2 𝑝 − 𝑞 5 = 144 9 + 2 ( 3 ) 5 ( )+ 5 ( ) 2 𝑝 − 𝑞 5 = 144 9 + 6 5+ 5 𝑝 − 𝑞 5 = 144 14 + 6 5 𝑝 − 𝑞 5 = 72 7 + 3 5 ★ Rationalise the denominator 𝑝 − 𝑞 5 = 72 7 + 3 5 × 7 − 3 57 − 3 5 𝑝 − 𝑞 5 = 504 − 216 549 − 9 ( 5 ) 𝑝 − 𝑞 5 = 504 − 216 54 𝑝 − 𝑞 5= 126 − 54 5 𝑝 = 126 , 𝑞 = 54 [6] 4. Simplify . 7 2− 3 8+ 32 7 2− 3 8+ 32 = 7 2− 3 4 × 2+ 2 × 16 = 7 2− 6 2+ 4 2 = 5 2 [2]

5. A rectangle has an area of . Given that the length is cm, 11 − 7 𝑐 𝑚 2 3 + 7 find the breadth of the rectangle and express the breadth in the form , 𝑎 − 𝑏 7 where and are integers. 𝑎 𝑏 Area of rectangle = 𝑙𝑏 11 − 7= 3 + 7 ( )𝑏 11 − 73 + 7 = 𝑏 ★ Rationalise the denominator 𝑏 = 11 − 73 + 7 × 3 − 73 − 7 𝑏 = 33 − 11 7− 3 7+ 7 9 − 7 𝑏 = 40 − 14 72 𝑏 = 20 − 7 7 cm 𝑏 = 20 − 7 7 [4] 6. It is given that . Find in the form , 2 3( 𝑥 + 1 ) = 2 − 𝑥 𝑥 𝑎 3− 𝑏 where and are real numbers. 𝑎 𝑏 2 3( 𝑥 + 1 ) = 2 − 𝑥 2 3𝑥 + 2 3= 2 − 𝑥 2 3𝑥 + 𝑥 = 2 − 2 3 𝑥 2 3+ 1 ( )= 2 − 2 3 𝑥 = 2 − 2 32 3+ 1 ★ Rationalise the denominator 𝑥 = 2 − 2 32 3+ 1 × 2 3− 1 2 3− 1 𝑥 = 4 3− 2 − 4 ( 3 ) + 2 34 ( 3 ) − 1 𝑥 = 6 3− 14 11 𝑥 = 6 11 3 − 14 11 [4]

7. Show can be written in the form , where and 2 5− 3 ( ) 6 + 5 ( )3 + 5 𝑎 + 𝑏 5 𝑎 𝑏 are rational numbers. 2 5− 3 ( ) 6 + 5 ( )3 + 5 ★ Rationalise the denominator = 12 5+ 2 ( 5 ) − 18 − 3 53 + 5 × 3 − 53 − 5 = 9 5− 8 3 + 5 × 3 − 53 − 5 = 27 5− 9 ( 5 ) − 24 + 8 59 − 5 = 35 5− 69 4 = 35 54 − 69 4 = − 69 4 + 35 54 [4] 8. Solve the equation , where , giving your answer in its 𝑥 3 − 4 𝑥 = 3 𝑥 ≠ 0 simplest surd form. 𝑥 2 − 12 3 𝑥 = 3 𝑥 2 − 12 = 3 3𝑥 𝑥 2 − 3 3𝑥 − 12 = 0 𝑥 = − 𝑏 ± 𝑏 2 − 4 𝑎𝑐2 𝑎 𝑥 = − − 3 3 ( )± 3 3 ( ) 2 − 4 ( 1 ) ( − 12 )2 ( 1 ) 𝑥 = 3 3± 27 + 482 𝑥 = 3 3± 752 𝑥 = 3 3± 25 × 32 𝑥 = 3 3± 5 32 ot 𝑥 = 4 3 𝑥 = − 3 [4]

9. (a) Without using a calculator, express in the form , 2 3+ 2 ( ) 2 𝑎 + 𝑏 6 where and are integers. 𝑎 𝑏 2 3+ 2 ( ) 2 = 4 ( 3 ) + 2 2 3 ( ) 2 ( )+ 2 = 14 + 4 6 [2] 9. (b) Hence, or otherwise, express in the form , 2 3+ 2 ( ) 2 3 + 6 𝑝 + 𝑞 6 where and are rational numbers. 𝑝 𝑞 ★ Rationalise the denominator 14 + 4 63 + 6 × 3 − 63 − 6 = 42 − 14 6+ 12 6− 4 ( 6 ) 9 − 6 = 18 − 2 63 = 6 − 2 3 6 [3] 10. It is given that . Find in the form , where 3( 𝑥 + 3 ) = 𝑥 + 17 𝑥 𝑎 + 𝑏 3 where and are integers. 𝑎 𝑏 3( 𝑥 + 3 ) = 𝑥 + 17 3𝑥 + 3 3= 𝑥 + 17 3𝑥 − 𝑥 = 17 − 3 3 𝑥 3− 1 ( )= 17 − 3 3 ★ Rationalise the denominator 𝑥 = 17 − 3 33− 1 × 3+ 1 3+ 1 𝑥 = 17 3+ 17 − 3 ( 3 ) − 3 33 − 1 𝑥 = 14 3+ 8 2 𝑥 = 7 3+ 4 𝑥 = 4 + 7 3 [4]

11. Given that , find the values of the integers 𝑎 − 3 5 ( ) 2 + 5 ( )= 𝑏 − 2 5 and . 𝑎 𝑏 𝑎 − 3 5 ( ) 2 + 5 ( )= 𝑏 − 2 5 2 𝑎 + 𝑎 5− 6 5− 3 ( 5 ) = 2 𝑎 − 15 + 𝑎 5− 6 5By comparison, 𝑎 5− 6 5= − 2 5 𝑎 5= 4 5 𝑎 = 4 𝑏 = 2 ( 4 ) − 15 𝑏 = − 7 [3] 12. Solve , leaving your answer in the form , 𝑥 + 7− 3 2= 1 𝑚 + 𝑛 2 where and are integers. 𝑚 𝑛 𝑥 + 7− 3 2= 1 𝑥 + 7= 1 + 3 2 𝑥 + 7 = 1 + 3 2 ( ) 2 𝑥 + 7 = 1 + 2 ( 1 ) 3 2 ( )+ 9 ( 2 ) 𝑥 + 7 = 1 + 6 2+ 18 𝑥 = 19 − 7 + 6 2 𝑥 = 12 + 6 2 [3]

13. Simplify , giving your answer in the form , 108 − 12 3 𝑘 3 where is an integer. 𝑘 108 − 12 3 = 3 × 36 − 12 3 ★ Rationalise the denominator = 6 3 − 12 3 × 33 =6 3 − 12 33 = 18 3− 12 33 = 6 33